Multiples of 1 + X and alternating partial sums #
For a finite sequence d₀, …, d_N in a ring, write
qₙ = dₙ - dₙ₋₁ + ⋯ ± d₀ = ∑ᵢ₌₀ⁿ (-1)ⁿ⁻ⁱ dᵢ for its alternating partial sums. Telescoping gives
the decomposition ∑ᵢ₌₀ᴺ dᵢ Xⁱ = (1 + X) ∑ₙ₌₀ᴺ⁻¹ qₙ Xⁿ + q_N X^N
(Polynomial.sum_range_monomial_eq_one_add_X_mul_add_monomial). So 1 + X divides the polynomial
exactly when q_N = 0, and then the quotient is ∑ₙ₌₀ᴺ⁻¹ qₙ Xⁿ
(Polynomial.sum_range_monomial_eq_one_add_X_mul_iff).
Over ℕ this characterizes the numerical shape of the Morse inequalities. For sequences
b, c : ℕ → ℕ, there is a polynomial Q with natural coefficients such that
∑ᵢ₌₀ᴺ cᵢ Xⁱ = ∑ᵢ₌₀ᴺ bᵢ Xⁱ + (1 + X) Q if and only if the alternating partial sums satisfy
bₙ - bₙ₋₁ + ⋯ ± b₀ ≤ cₙ - cₙ₋₁ + ⋯ ± c₀ for n < N, with equality for n = N
(Polynomial.exists_sum_range_monomial_eq_add_one_add_X_mul_iff). If moreover bᵢ = cᵢ for
i > N, then the coefficient of Xⁿ in Q is the difference of the two alternating partial sums
in degree n, for every n (Polynomial.coeff_eq_sum_range_neg_one_pow_mul_sub).
References #
- J. Milnor, Morse Theory, Annals of Mathematics Studies 51, Princeton University Press, 1963, Section 5.
Telescoping against 1 + X. ∑ᵢ₌₀ᴺ dᵢ Xⁱ = (1 + X) ∑ₙ₌₀ᴺ⁻¹ qₙ Xⁿ + q_N X^N, where
qₙ = ∑ᵢ₌₀ⁿ (-1)ⁿ⁻ⁱ dᵢ. This is not division with remainder by the monic 1 + X (whose
remainder is the constant (-1)ᴺ q_N), but it shows that 1 + X divides the left side when
q_N = 0.
1 + X divides ∑ᵢ₌₀ᴺ dᵢ Xⁱ, with quotient Q, exactly when the alternating sum
q_N = ∑ᵢ₌₀ᴺ (-1)ᴺ⁻ⁱ dᵢ vanishes and Q = ∑ₙ₌₀ᴺ⁻¹ qₙ Xⁿ.
Polynomial form of the Morse inequalities. For b, c : ℕ → ℕ, there is a polynomial
Q ∈ ℕ[X] with ∑ᵢ₌₀ᴺ cᵢ Xⁱ = ∑ᵢ₌₀ᴺ bᵢ Xⁱ + (1 + X) Q if and only if
bₙ - bₙ₋₁ + ⋯ ± b₀ ≤ cₙ - cₙ₋₁ + ⋯ ± c₀ for every n < N, with equality for n = N.
If ∑ᵢ₌₀ᴺ cᵢ Xⁱ = ∑ᵢ₌₀ᴺ bᵢ Xⁱ + (1 + X) Q in ℕ[X], and bᵢ = cᵢ for i > N, then the
coefficient of Xⁿ in Q is (cₙ - cₙ₋₁ + ⋯ ± c₀) - (bₙ - bₙ₋₁ + ⋯ ± b₀), for every n.